Derivation of the Spin Memory Effect
The Observable: Velocity Circulation
Place test particles on a closed curve $\mathcal{C}$ on a 2-sphere of radius $r \gg GM/c^2$. Define the velocity circulation:
$$\Gamma(\mathcal{C}) = \oint_{\mathcal{C}} \dot{x}^A\,dl_A$$
After a GW burst, $\Gamma$ acquires a permanent offset $\Delta\Gamma$ — the spin memory. It detects the curl (magnetic parity), not displacement (electric parity).
Transverse Velocity from the Geodesic Equation
At $\mathcal{O}(1/r)$, the geodesic equation gives:
$$\partial_u v_A = \tfrac{1}{2}N_{AB}\,v^B + \partial_u\left(g_{uA}\big|_{\mathcal{O}(r^{-1})}\right) + \mathcal{O}(r^{-2})$$
Integrating over the full burst:
$$v_A(+\infty) = \tfrac{1}{2}\int_{-\infty}^{+\infty} N_{AB}\,v^B\,du + \Delta\left(g_{uA}\big|_{1/r}\right)$$
Isolating the Magnetic-Parity Part
Applying the curl operator $\epsilon^{AB}D_A$ to isolate the magnetic-parity (odd) component:
$$\epsilon^{AB}D_A v_B\big|_{+\infty} = \tfrac{2}{3}\,\epsilon^{AB}D_A\Delta N_B$$
The Spin Memory Formula
$$\boxed{\Delta\Gamma(\mathcal{C}) = \frac{2}{3r}\oint_{\mathcal{C}}\Delta N_A\,dx^A = \frac{2}{3r}\int_\Sigma \epsilon^{AB}D_A\Delta N_B\,d^2\Sigma}$$
Permanent Change in the Angular Momentum Aspect
$$\Delta N_A = \int_{-\infty}^{+\infty}\partial_u N_A\,du = \frac{1}{4}\int_{-\infty}^{+\infty}\left[\epsilon_{AB}D^B\left(N_{CD}N^{CD}\right) + 2\,N_{BC}\,D^B C^C{}_A\right]du + \Delta(\text{shear boundary})$$
Note: The leading vacuum contribution is sourced by $N_{CD}N^{CD}$ — the same GW energy flux that drives Christodoulou displacement memory — but projected onto odd parity via $\epsilon_{AB}$.
Displacement Memory vs Spin Memory
$$\underbrace{\Delta\xi^i \sim \Delta C_{AB}}_{\text{even parity, }E\text{-mode}} \longleftrightarrow \underbrace{\Delta\Gamma \sim \epsilon^{AB}D_A\Delta N_B}_{\text{odd parity, }B\text{-mode}}$$